Question 1015841
Let's look at the coefficient matrix,
{{{A=(matrix(2,2,
3,m,
m+2,5))}}}
The determinant is then,
{{{15-m(m+2)=15-m^2-2m=-m^2-2m+15}}}
Look for values of {{{m}}} that make the determinant equal to zero,
{{{m^2-2m+15=0}}}
{{{(m+5)(m-3)=0}}}
So then when {{{m=-5}}}, the equations become,
{{{3x-5y=5}}}
{{{-3x+5=-5}}}
and you see that the second equation is just the first equation multiplied by {{{-1}}}.
So this dependent system has infinitely many solutions.
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When {{{m=3}}}, the equations become,
{{{3x+3y=5}}} or {{{x+y=5/3}}}
{{{5x+5y=3}}} or {{{x+y=3/5}}}
So the equations are now parallel and therefore have no solution.